What is the tangent of 1/(2sqrt(x+1)) ?

The tangent of 1/(2sqrt(x+1)) is y=-1/(4(a_{0)+1)^{3/2}}x+(3a_{0}+2)/(4(a_{0)+1)^{3/2}}

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tangent of 1/(2sqrt(x+1))

tangent of

\frac{1}{2 x + 1}

y = - \frac{1}{4 ( a _{0} + 1 ) ^{\frac{3}{2}}} x + \frac{3 a _{0} + 2}{4 ( a _{0} + 1 ) ^{\frac{3}{2}}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{1}{2 a _{0} + 1})

Find the slope of

y = \frac{1}{2 x + 1} : \frac{d y}{d x} = - \frac{1}{4 ( x + 1 ) ^{\frac{3}{2}}}

Compute the slope of

y = \frac{1}{2 x + 1}

x = a_{0} : m = - \frac{1}{4 ( a _{0} + 1 ) ^{\frac{3}{2}}}

Find the line with slope m=

- \frac{1}{4 ( a _{0} + 1 ) ^{\frac{3}{2}}}

and passing through

(a_{0}, \frac{1}{2 a _{0} + 1}) : y = - \frac{1}{4 ( a _{0} + 1 ) ^{\frac{3}{2}}} x + \frac{3 a _{0} + 2}{4 ( a _{0} + 1 ) ^{\frac{3}{2}}}

y = - \frac{1}{4 ( a _{0} + 1 ) ^{\frac{3}{2}}} x + \frac{3 a _{0} + 2}{4 ( a _{0} + 1 ) ^{\frac{3}{2}}}

\int \frac{2 x}{( 2 x ^{2} - 3 ) ^{3}} d x

\int arctan (4 t) d t

y = 0, y = 2 x^{2}, y = 0, [0, 1]

n = 3 \sum \infty \frac{1}{ln ( n )}

\int \frac{1}{( x - 5 ) ( x - 2 )} d x

What is the tangent of 1/(2sqrt(x+1)) ?
The tangent of 1/(2sqrt(x+1)) is y=-1/(4(a_{0)+1)^{3/2}}x+(3a_{0}+2)/(4(a_{0)+1)^{3/2}}